A system is taken from state A to state B along two different paths $1$ and $2$ $.$ The work done on the system along these two paths are $W_{1}$ and $W_{2}$ respectively $.$ The heat absorbed by the system along these two paths are $Q_{1}$ and $Q_{2}$ respectively. The internal energies at A and B are $U_{A}$ and $U_{B}$ respectively. Then,

W_{1} = W_{2} = U_{B} - U_{A}

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Q_{1} = Q_{2} = U_{A} - U_{B}

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Q_{1} + W_{1} = Q_{2} + W_{2} = U_{A} + U_{B}

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Q_{1} + (- W_{1}) = Q_{2} + (W_{2}) = U_{B} - U_{A}

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Solution

The correct option is A $W_{1} = W_{2} = U_{B} - U_{A}$
Since internal energy is state function i.e.
it does not depend on the path.
So, change in internal is for path $1$ and $2$ will be.
$Δ U = U_{B} - U_{B}$
Now,
from $I$ law of thermodynamies.
for path $1,$
$Q_{1} = W_{1} + Δ U$
$\Rightarrow Δ U = Q_{1} - W_{1}$ ......(i)
for path 2
$Q_{2} = W_{2} + Δ U$
$\Rightarrow Δ U = Q_{2} - W_{2}$ ....(ii)
from $(i)$ & $(i i)$
$Δ U = Q_{1} - W_{1} = Q_{2} - W_{2}$
$Q_{1} - W_{1} = Q_{2} - W_{2} = U_{B} - U_{A}$

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